Question

When mean optical power launched into an 8 km length of fiber is 12 \(\mu\)W, the mean optical power at the fiber output is 3 \(\mu\)W. Find the overall signal attenuation in dB

• A. 15 dB
• B. 16 dB
• C. 10 dB
• D. 12 dB

Hint:

Attenuation is measured as \(10log_{10}(P_{out}/P_{in})\) , pay close attention to the unit of measure which is decibels (dB).

Answer 1

The Correct Option is A

Step 1: The power launched is \( P_{in} = 12 \mu W \) and the power at the output is \( P_{out} = 3 \mu W \). The attenuation in dB is calculated as: \[ Attenuation(dB) = 10 \log_{10} \frac{P_{out}}{P_{in}} = 10 \log_{10} \frac{3}{12} \] \[ = 10 \log_{10} 0.25 = 10 \times (-0.602) = -6.02 dB \] The negative sign indicates the loss in the fiber. 
Step 2: Since attenuation is a measure of loss of power, we take the absolute value. Therefore, the overall signal attenuation is 6 dB.